Question: The equation of a circle $C$ is $x^2+y^2+6x-8y-11 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+6x) + (y^2-8y) = 11$ $(x^2+6x+9) + (y^2-8y+16) = 11 + 9 + 16$ $(x+3)^{2} + (y-4)^{2} = 36 = 6^2$ Thus, $(h, k) = (-3, 4)$ and $r = 6$.